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malfutka [58]
3 years ago
14

The amount of time that customers wait in line during peak hours at one bank is normally distributed with a mean of 13 minutes a

nd a standard deviation of 3 minutes. The percentage of time that the waiting time lies between 11 and 13 minutes is equal to the area under the standard normal curve between ___and ___.
a).62, .77
b) -.67,0
c)-2, 0
d) 0, .67
Mathematics
1 answer:
Neporo4naja [7]3 years ago
8 0

Answer: I believe the answer is c)-2, 0

Step-by-step explanation: The amount of time that customers wait in line during peak hours at one bank is normally distributed with a mean of 13 minutes and a standard deviation of 3 minutes then the answer should be c)-2, 0 ?

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Elias has decided to fence in a garden that is in the shape of a parallelogram. If one side measures 76 ft and 250 ft of fencing
Leni [432]

The length of side of garden are 76 feet and 49 feet

<em><u>Solution:</u></em>

Given that, Elias has decided to fence in a garden that is in the shape of a parallelogram

Measure of one side = 76 feet

250 ft of fencing is needed to enclose the garden

Therefore, perimeter = 250

<em><u>The perimeter of parallelogram is given by:</u></em>

Perimeter = 2(a + b)

Where, a and b are the length of sides

Here, a = 76

b = ?

<em><u>Substituting in formula, we get</u></em>

250 = 2(76 + b)

250 = 152 + b

2b = 250 - 152

2b = 98

b = 49

Thus the length of side of garden is 49 feet

8 0
2 years ago
the numbers 1,2,3,4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacet. find t
Tresset [83]

Consider such events:

A - slip with number 3 is chosen;

B - the sum of numbers is 4.

You have to count Pr(A|B).

Use formula for conditional probability:

Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}.

1. The event A\cap B consists in selecting two slips, first is 3 and second should be 1, because the sum is 4. The number of favorable outcomes is exactly 1 and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event A\cap B is

Pr(A\cap B)=\dfrac{1}{20}.

2. The event B consists in selecting two slips with the sum 4. The number of favorable outcomes is exactly 2 (1st slip 3 and 2nd slip 1 or 1st slip 1 and 2nd slip 3) and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event B is

Pr(B)=\dfrac{2}{20}=\dfrac{1}{10}.

3. Then

Pr(A|B)=\dfrac{\frac{1}{20} }{\frac{1}{10} }=\dfrac{1}{2}.

Answer: \dfrac{1}{2}.

5 0
3 years ago
In Ms. Jordan's math class,
vovangra [49]

Answer:

there are 27 female

Step-by-step explanation:

3 0
3 years ago
⦟KLJ and ⦟JLM are a linear pair. Find the measures of ⦟KLJ and ⦟JLM.
KATRIN_1 [288]

Answer:

Step-by-step explanation:

6x + 12 + 4x + 8 =  180 {Linear pair}

6x + 4x + 12 + 8 = 180 {Combine like terms}

         10x + 20  = 180   {Subtract 20 from both sides}

              10x =  180 - 20

             10x  = 160    {Divide both sides by 10}

               x = 160/10

                x = 16

m∠KLJ = 6*16 + 12

            = 96 +12

            = 108

m∠JLM = 4*16 + 8

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8 0
2 years ago
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Solve the following quadratic-linear system of equations.
Reika [66]
The answer should and must be Done

7 0
3 years ago
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